2
$\begingroup$

Okay so I got this question in a rehearsal test today , and got completely stumped.

Let $f$ be a differentiable function satisfying the functional rule $f(xy)=f(x)+f(y)+\frac{x+y-1}{xy}$, $\forall$ $x,y>0$ and $f'(1)=2$.The question asked some more questions regarding $f(x)$,but I am unable to find $f(x)$.

My try:I tried to treat $y$ as a constant and then differentiated both sides of the equation,making use of the fact that $f'(1)=2$ ; but that got me nowhere.Someone kindly help.

Thanks in advance.

$\endgroup$
  • $\begingroup$ try substituting y=1/x and the differentiate $\endgroup$ – Savitar May 10 '17 at 15:10
  • $\begingroup$ How does that help ? $\endgroup$ – Zlatan May 10 '17 at 15:41
5
$\begingroup$

Your try gives

$$yf'(xy) = f'(x) -\frac{1}{x^2} + \frac{1}{x^2y},\tag{1}$$

for all $x,y > 0$. Choosing $x = 1$ in $(1)$, we obtain

$$yf'(y) = 2 - 1 + \frac{1}{y},$$

or

$$f'(y) = \frac{1}{y} + \frac{1}{y^2}.\tag{2}$$

From that, we can easily determine $f$ up to addition of a constant, and we can determine the constant from the rule e.g. by setting $x = y = 1$ there.

An alternative way to find the function is to rewrite the functional equation as

$$f(xy) + \frac{1}{xy} = \biggl(f(x) + \frac{1}{x}\biggr) + \biggl(f(y) + \frac{1}{y}\biggr),$$

and if one knows the continuous functions satisfying the functional equation $g(xy) = g(x) + g(y)$, $f$ is easily determined.

$\endgroup$
  • $\begingroup$ Sorry , Thanks for your effort into this , but like a noob I am ,Im not getting anything.How do i get $f(x)$ from this ? And how do I tackle similar problems?Isnt there a thumb rule? $\endgroup$ – Zlatan May 10 '17 at 15:50
  • $\begingroup$ We have a nice form for the derivative of $f$. One can read the primitives of $f'$ off that if one knows the derivatives of the most famous functions. If $g'(y) = y^{\alpha}$, what is $g(y)$? $\endgroup$ – Daniel Fischer May 10 '17 at 16:01
1
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{f}}$ is differentiable. $\ds{\mrm{f}\pars{xy} = \,\mrm{f}\pars{x} + \,\mrm{f}\pars{y} + {1 \over y} + {1 \over x} - {1 \over xy}\,,\quad x, y \in \mathbb{R}_{\ >\ 0}\quad\mbox{and}\quad\mrm{f}'\pars{1} = 2}$.

$$ \bbx{\quad\mbox{Note that}\quad\mrm{f}\pars{1} = -1\quad \pars{~\mbox{from the 'original functional equation'}~}\quad} $$

  • Derive both members, of the above functional equation, respect of $\ds{x}$: \begin{equation} \mrm{f}'\pars{xy}y = \,\mrm{f}'\pars{x} - {1 \over x^{2}} + {1 \over x^{2}y} \implies \mrm{f}'\pars{xy}xy = \,\mrm{f}'\pars{x}x - {1 \over x} + {1 \over xy} \label{1}\tag{1} \end{equation}
  • Derive both members, of the above functional equation, respect of $\ds{y}$: \begin{equation} \mrm{f}'\pars{xy}x = \,\mrm{f}'\pars{y} - {1 \over y^{2}} + {1 \over xy^{2}} \implies \mrm{f}'\pars{xy}xy = \,\mrm{f}'\pars{y}y - {1 \over y} + {1 \over xy} \label{2}\tag{2} \end{equation} \eqref{1} and \eqref{2} lead to: \begin{align} &\,\mrm{f}'\pars{x}x - {1 \over x} = \,\mrm{f}'\pars{y}y - {1 \over y} = \alpha\,, \qquad\pars{~\alpha:\ \mbox{a}\ x\ \mbox{and}\ y\ \mbox{independent constant}~} \end{align}

    $\ds{\alpha = \,\mrm{f}'\pars{1} \times 1 - {1 \over 1} = 1}$.

Now, you are left with $$ \,\mrm{f}'\pars{x} = {1 \over x} + {1 \over x^{2}}\,,\quad\mrm{f}\pars{1} = -1 \implies \bbox[#ffe,10px,border:1px dotted navy]{\ds{\mrm{f}\pars{x} = \ln\pars{x} - {1 \over x}}} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.